Euler Class of a vector field Let M be a closed 3-manifold and let X be a vector field on M. In which conditions might we define a Euler class associated to X? For example, is it possible to define for a rotational Beltrami fields?
I have been studying a Etnyre and Ghrist's paper (Contact Topology and Hydrodynamics) where they proof that every rotational Beltrami fields on a 3-manifold M is a Reeb-like field for some contact form on M.
 A: There is a refined notion of an Euler class that takes vector fields into account, and this refined notion captures some nontrivial information about vector fields.
Let me phrase the definition in a more general context. Assume that $V \to X$ is a rank real oriented vector bundle on a space. Let $\tau \in H^n (V, V_0)$ be the Thom class ($V_0$ is the complement of the zero section). Then it is well-known that the Euler class of $V$ is the pullback of $\tau$ through the zero section of $V$; and any other section will give the same result.
Moreover, if $V$ is odd-dimensional, then the Euler class is $2$-torsion, since the antipodal map of $V$ defines an isomorphism $V \cong V^{op}$ (with the opposite orientation). 
The refined version of the Euler class can be defined if a section $s: X \to V$ is given which is nonzero on the (say closed) subspace $Y \subset X$. In that case, define
$$e(V,s) := s^{*} \tau \in H^n (X,Y).$$
Under the map $H^* (X,Y) \to H^* (X)$, this relative Euler class maps to the usual Euler class. The Euler class $E(V,s)$ only depends on the homotopy class of sections, where ''homotopy'' means homotopy through sections that are nonzero on $Y$. 
Note that this Euler class does not need to be $2$-torsion when the vector bundle is odd-dimensional: the antipodal map does not fix the section except where it is zero.
Here is an interesting example. Let $M$ be a manifold with boundary, and let $s$ be a vector field which has no zeroes on $\partial M$. The relative Euler class $e(TM,s) \in H^n (M, \partial M)$ is an obstruction to extend the vector field $s|_{\partial M}$ to all of $M$ without zeroes (and under the assumption: $\pi_0 (\partial M) \to \pi_0 (M)$ surjective, it is the only obstruction). 
There are more interesting aspects, and I can give you more answers if your question is more specific.
